Question

Silver phosphate, Ag3PO4, is an insoluble salt that has a Ksp = 1.3 x 10-20.

a. Calculate the molar solubility of Ag3PO4 in pure water.

b. Will the molar solubility of Ag3PO4 increase or decrease in a solution containing 0.020 M Na3PO4?

c. Will a precipitate form if a solution is made by combining 3.0 x 10-5 M AgNO3 and 2.3 x 10-4 M Na3PO4? (hint: calculate Q)

Answer 1

a)

Consider reaction, Ag3PO4 (s) $\rightleftharpoons$ 3 Ag + (aq) + PO 4 3- (aq)

K sp = [ Ag + ] 3 [ PO 4 3- ] = 1.3 $\times$ 10 -20

Let 'S' mol / L be the solubility of Ag3PO4 in pure water , then [ Ag + ] = ' 3 S ' mol / L and [ PO 4 3- ] = 'S' mol / L.

$\therefore$ K sp = [ Ag + ] 3 [ PO 4 3- ] = 1.3 $\times$ 10 -20  = ( 3 S ) 3 S

27 S 3  $\times$ S = 1.3 $\times$ 10 -20

27 S 4 =  1.3 $\times$ 10 -20  

S 4 =  1.3 $\times$ 10 -20  / 27

S 4 = 4.81  $\times$ 10 -22

S = 4.7 $\times$ 10 -06 M

ANSWER :  solubility of Ag3PO4 in pure water = 4.7 $\times$ 10 -06 M

b)

Consider reaction, Ag3PO4 (s) $\rightleftharpoons$ 3 Ag + (aq) + PO 4 3- (aq)

According to Le-chateliars principle, addition of either Ag + or PO 4 3- will shift reaction to the left side.

If we add 0.02 M PO 4 3- solution to above equilbrium reaction, reaction will proceed to produce Ag3PO4 (s) . Therefore, addition of 0.02 M PO 4 3- will decrease the solubility of Ag3PO4 .

ANSWER : Solubility decreases.

C)

We know that precipitation takes place when Ionic product > solubility product.

To check precipitation , we need to calculate ionic product.

Let's calculate Ionic product.

We have, Q = [ Ag + ] 3 [ PO 4 3- ]

Substituting given values in above equation , we get

Q = ( 3.0 $\times$ 10 -05 ) 3  $\times$ ( 2.3 $\times$ 10 -04 )

Q = 6.21 $\times$ 10 -18

Ionic product > Solubility product , hence precipitation will be formed in the reaction.

ANSWER : Yes, ppt. will be formed.